| Exam Board | OCR |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2013 |
| Session | January |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Substitution to find new equation |
| Difficulty | Moderate -0.3 This is a straightforward FP1 question on root transformations requiring basic substitution and application of Vieta's formulas. Part (i) is mechanical algebra, and part (ii) follows directly from recognizing the transformed roots. While it requires understanding of root relationships, it's a standard textbook exercise with no novel insight needed, making it slightly easier than average. |
| Spec | 4.05a Roots and coefficients: symmetric functions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer: \(4u^2+6u+k+2=0\) | M1, A1 [2] | Substitute and attempt to simplify; Obtain correct answer, must be an equation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer: Either \(\frac{k+2}{4}\) or \(\frac{k+2}{4}\) | M1, A1ft; M1, A1 [2] | Use products of roots of new quadratic i.e. use \((\pm) c/a\); Obtain correct answer, from their quadratic; Use sum and product of roots of original equation; Obtain correct answer |
### (i)
Answer: $4u^2+6u+k+2=0$ | M1, A1 [2] | Substitute and attempt to simplify; Obtain correct answer, must be an equation
### (ii)
Answer: Either $\frac{k+2}{4}$ or $\frac{k+2}{4}$ | M1, A1ft; M1, A1 [2] | Use products of roots of new quadratic i.e. use $(\pm) c/a$; Obtain correct answer, from their quadratic; Use sum and product of roots of original equation; Obtain correct answerThe quadratic equation $x^2 + x + k = 0$ has roots $\alpha$ and $\beta$.
\begin{enumerate}[label=(\roman*)]
\item Use the substitution $x = 2u + 1$ to obtain a quadratic equation in $u$. [2]
\item Hence, or otherwise, find the value of $\left(\frac{\alpha - 1}{2}\right)\left(\frac{\beta - 1}{2}\right)$ in terms of $k$. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR FP1 2013 Q4 [4]}}